George, I see where he says 'If we assume that the result proved for a polygon...'. In the preceeding paragraph he says ' It is at once apparent that this result still holds good if the clock moves from A to B in any polygonal line, ...'. But it isn't apparent to me, so by way of trying to understand, I will make something up. Check this out.
Let A be at x,y,z = 0 and B be at 2,0,0 in the K frame. A clock moving along the x-axis will follow the usual slow clock formula developed in his previous paragraphs. Now let the polygon consist of the line from x,y = 0,0 to x,y = 1,1 , then to 2,0, then to 1, -1, and ending at 0,0, all in the x-y plane. Introduce 'local' coordinate frames with their x axes aligned with each line segment. Let one local frame be at rest in K, and the other move along each segment starting at 0,0 and moving in a clockwise direction. Applying the slow clock formula to each segment in turn and adding gives the total time delay on arrival back at A. The time from A to B along the x-axis will be different from the time along the polygon, but the same formula 'holds good'.
Thats how his analysis makes sense to me. No use is made of acceleration. That suggests that the analysis is purely geometrical. The time delay around the polygon will be the same as along a straight line whose length equals the perimeter of the polygon. I sense that this example is out of context and that further study might provide meaningful explanations.
JM